Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $p = \dfrac{5z^2 + 20z - 160}{-5z^2 - 5z + 100} \div \dfrac{-4z^2 - 4z}{2z^2 + 10z} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{5z^2 + 20z - 160}{-5z^2 - 5z + 100} \times \dfrac{2z^2 + 10z}{-4z^2 - 4z} $ First factor out any common factors. $p = \dfrac{5(z^2 + 4z - 32)}{-5(z^2 + z - 20)} \times \dfrac{2z(z + 5)}{-4z(z + 1)} $ Then factor the quadratic expressions. $p = \dfrac {5(z - 4)(z + 8)} {-5(z - 4)(z + 5)} \times \dfrac {2z(z + 5)} {-4z(z + 1)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac { 5(z - 4)(z + 8) \times 2z(z + 5)} { -5(z - 4)(z + 5) \times -4z(z + 1)} $ $p = \dfrac {10z(z - 4)(z + 8)(z + 5)} {20z(z - 4)(z + 5)(z + 1)} $ Notice that $(z - 4)$ and $(z + 5)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac {10z\cancel{(z - 4)}(z + 8)(z + 5)} {20z\cancel{(z - 4)}(z + 5)(z + 1)} $ We are dividing by $z - 4$ , so $z - 4 \neq 0$ Therefore, $z \neq 4$ $p = \dfrac {10z\cancel{(z - 4)}(z + 8)\cancel{(z + 5)}} {20z\cancel{(z - 4)}\cancel{(z + 5)}(z + 1)} $ We are dividing by $z + 5$ , so $z + 5 \neq 0$ Therefore, $z \neq -5$ $p = \dfrac {10z(z + 8)} {20z(z + 1)} $ $ p = \dfrac{z + 8}{2(z + 1)}; z \neq 4; z \neq -5 $